Answer to Question #160005 in Quantum Mechanics for Bhavana

Question

Calculate de broglie wavelength and velocity of an electron carrying kinetic energy 2keV

Answer 1

From the definition of the de Broglie wavelength, we have:

$$\lambda=\dfrac{h}{p}=\dfrac{h}{mv}.$$

Let's find the velocity of the electron from the definition of the kinetic energy:

$$KE=\dfrac{1}{2}mv^2,$$ $$v=\sqrt{\dfrac{2KE}{m}}.$$

Substituting $v$ into the first equation, we get:

$$\lambda=\dfrac{h}{\sqrt{2mKE}},$$

$\lambda=\dfrac{6.626\cdot10^{-34}\ J\cdot s}{\sqrt{2\cdot9.1\cdot10^{-31}\ kg\cdot2\cdot10^3\ eV\cdot1.6\cdot10^{-19}\ \dfrac{J}{eV}}}=2.74\cdot10^{-11}\ m.$

Finally, we can find the velocity of the electron:

$$v=\sqrt{\dfrac{2\cdot2\cdot10^3\ eV\cdot1.6\cdot10^{-19}\ \dfrac{J}{eV}}{9.1\cdot10^{-31}\ kg}}=2.65\cdot10^7\ \dfrac{m}{s}.$$